Sequential spaces, V

In the previous post Sequential spaces, IV, we show that the uncountable product of sequential spaces is not sequential (e.g. the product $2^{\omega_1}$ is not sequential). What is more remarkable is that the product of two sequential spaces needs not be sequential. We present an example of a first countable space and a Frechet space whose product is not a k-space (thus not sequential). For the previous discussion on this blog on sequential spaces and k-spaces, see the links at the end of this post.

Let $\mathbb{R}$ be the real line and let $\mathbb{N}$ be the set of all positive integers. Let $X$ be the space $\mathbb{R}-\left\{1,\frac{1}{2},\frac{1}{3},\cdots\right\}$ with the topology inherited from the usual topology on the real line. Let $Y=\mathbb{R}$ with the positive integers identified as one point (call this point $p$). We claim that $X \times Y$ is not a k-space and thus not a sequential space. To this end, we define a non-closed $A \subset X \times Y$ such that $K \cap A$ is closed in $K$ for all compact $K \subset X \times Y$.

Let $A=\bigcup \limits_{i=1}^\infty A_i$ where for each $i \in \mathbb{N}$, the set $A_i$ is defined by the following:

$\displaystyle A_i =\left\{\biggl(\frac{1}{i}+\frac{a_i}{j},i+\frac{0.5}{j} \biggr) \in X \times Y:j \in \mathbb{N}\right\}$

where $\displaystyle a_i=\biggl(\frac{1}{i}-\frac{1}{i+1} \biggr) 10^{-i}$.

Clearly $A$ is not closed as $(0,p) \in \overline{A}-A$. In fact in the product space $X \times Y$, the point $(0,p)$ is the only limit point of the set $A$. Another observation is that for each $n \in \mathbb{N}$, $(0,p)$ is not a limit point of $\bigcup \limits_{i=1}^n A_i$. Furthermore, if $z_i \in A_i$ for each $i \in S$ where $S$ is an infinite subset of $\mathbb{N}$, then $(0,p)$ is not a limit point of $\left\{z_i:i \in S\right\}$. It follows that no infinite subset of $A$ is compact. Consequently, $K \cap A$ is finite for each compact $K \subset X \times Y$. Thus $X \times Y$ is not a k-space. To see that $X \times Y$ is not sequential directly, observe that $A$ is sequentially closed.

Previous posts on sequential spaces and k-spaces